What is the relationship between normality and molarity? Normality and molality? Mole fraction and molarity?
1 Answer
Make sure to read this answer carefully... these are very conditional relationships.
You have asked for:
#"N"# #-># #"M"# #"N"# #-># #"m"# #chi -> "m"# (I added this one)#chi -> "M"#
DISCLAIMER: LONG ANSWER!
NORMALITY TO MOLARITY
The hardest one to consider is
Then, normality is defined with respect to the
We define
#color(blue)(barul|stackrel(" ")(" ""N" = (c)/(f_(eq))" ")|)# ,
#c# in#"mol/L"#
Keep in mind that
NORMALITY TO MOLALITY
Well, this is not easy either. Normality is proportional to molarity
#c = "mols solute"/"L solution" = n_"solute"/V_"soln"#
#= color(green)((n_"solute")/(w_"soln") xx rho_"soln")# where
#rho# is the density in#"g/L"# and#w# is the mass in#"g"# .
#m = "mols solute"/"kg solvent" = n_"solute"/w_"solvent"#
#= "mols solute"/("kg solution - kg solute")#
#= color(green)(n_"solute"/(w_"soln" - w_"solute"))#
Since we already know how to get from normality to molarity... consider getting from molarity to molality:
#overbrace((n_"solute"rho_"soln")/(w_"soln"))^"Molarity" " vs. " overbrace(n_"solute"/(w_"soln" - w_"solute"))^"Molality"#
Only for sufficiently dilute solutions, i.e. the mass of solvent overwhelms the mass of solute, do we have
#c ~~ m xx rho_"solvent"#
This is usually considered to be around
#color(blue)(barul|stackrel(" ")(" ""N" ~~ (m xx rho_"solvent")/(1000 "g"/"kg" xx f_(eq))" ")|)# ,
#m# in#"mol/kg"#
#rho# in#"g/L"#
MOL FRACTION TO MOLALITY
For a mol fraction, consider a simple binary ideal mixture (one solute in one solvent):
#chi_"solute" = n_"solute"/(n_"soln") = n_"solute"/(n_"solute" + n_"solvent")# ,where
#n_"solute"# is the mols of solute and#n_"soln" = n_"solute" + n_"solvent"# .
Refer back to the definition of molality (not to be confused with mass!!) to find:
#m = n_"solute"/w_"solvent"# ,
The mols of solute can be found from the mol fraction.
#chi_"solute"(n_"solute" + n_"solvent") = n_"solute"#
#chi_"solute"n_"solute" + chi_"solute"n_"solvent" = n_"solute"#
#chi_"solute"n_"solvent" = (1 - chi_"solute")n_"solute"#
#n_"solute" = (chi_"solute"n_"solvent")/(1 - chi_"solute")#
Substituting back into the expression for the molality:
#m = (chi_"solute")/(1 - chi_"solute") xx n_"solvent"/w_"solvent"#
But the mols divided by mass is one over the molar mass,
#color(blue)(barul|stackrel(" ")(" "m = (1000 "g"/"kg")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" ")|)# ,
#M# in#"g/mol"#
#0 < chi < 1#
If you need it the other way around, you can solve for
MOL FRACTION TO MOLARITY
And for molarity, it is analogous, by adding in the result from earlier... which, again, we defined for sufficiently-dilute solutions!
#c ~~ m xx rho_"solvent"#
Thus, for solutions approximately
#color(blue)(barul|stackrel(" ")(" "c ~~ (rho_"solvent")/M_"solvent" xx (chi_"solute")/(1 - chi_"solute")" ")|)# ,
#M# in#"g/mol"#
#rho# in#"g/L"#
#0 < chi < 1#
For this relationship, you should find that